Results on Extremal Solutions for a Class of Boundary Value Problem of Nonlinear Fractional Order Differential Equations

Abstract

This paper investigates a class of boundary value problems involving Caputo fractional derivatives of order $\nu \in (2,3]$. We begin by establishing two novel comparison principles. Subsequently, by employing the monotone iterative technique coupled with upper and lower solutions, we demonstrate the existence of extremal solutions for the corresponding fractional differential equations. Finally, an illustrative example is provided to validate our main findings.

1. Introduction

In recent years, fractional differential equations (FDEs) have attracted significant attention due to their ability to model complex physical and engineering phenomena with memory effects, hereditary properties, and nonlocal dynamics. These include applications in viscoelastic materials [1], anomalous diffusion [2,3,4], and biological system modeling [5,6,7]. Beyond these established domains, the integration of fractional calculus with advanced nonlinear analytical frameworks is gaining unprecedented traction across a broader interdisciplinary spectrum. In particular, recent works have demonstrated the growing relevance of these combined approaches in addressing complex challenges in advanced control systems [8], neural networks [9], and uncertainty modeling [10]. Due to their profound significance and expanding applicability, the study of fractional differential equations has continued to attract much attention; we refer the reader to [11,12,13,14,15,16]. For example, in [13], Cabada and Wang considered the following fractional boundary value problems:

$$\left\{\begin{array}{c}{\displaystyle {}^{c}D_{0^{+}}^{\alpha }u\left(t\right)+f(t,u\left(t\right))=0,t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=u^{″}\left(0\right)=0,u\left(1\right)=\lambda {\int }_0^{1}u\left(s\right)ds,}\hfill \end{array}\right.$$

(1)

where $2<\alpha <3$, $0<\lambda <2$, $f:[0,1]\times [0,\infty )\to [0,\infty )$ is a continuous function and ${}^{C}D_{0^{+}}^{\alpha }$ is the Caputo fractional derivative. The authors found some existence results for positive solutions to (1) based on the known Guo–Krasnoselskii fixed point theorem.

In [14], Haddouchi investigated the following nonlocal fractional boundary value problem with a Riemann–Stieltjes integral boundary condition

$$\left\{\begin{array}{c}{\displaystyle {}^{c}D_{0^{+}}^{\alpha }u\left(t\right)+f(t,u\left(t\right))=0,t\in (0,1),}\hfill \\ {\displaystyle u\left(0\right)=u^{″}\left(0\right)=0,u\left(1\right)=\mu u\left(\eta \right)+\beta \gamma \left[u\right],}\hfill \end{array}\right.$$

where $2<\alpha \le 3$, $\eta \in (0,1)$, $\mu$, $\beta \ge 0$, $f:[0,1]\times [0,\infty )\to [0,\infty )$ is a continuous function, and $\gamma$ is a continuous linear functional given by the Riemann–Stieltjes function

$$\gamma \left[u\right]={\int }_0^{1}u\left(s\right)dA\left(s\right),$$

where $A:[0,1]\to \mathbb{R}$ is a bounded variation function. The author established the existence of positive solutions by means of the spectral analysis of the relevant linear operator and Gelfand’s formula.

However, the analytical solutions of nonlinear fractional order differential equations are usually difficult to express explicitly. Consequently, research efforts have focused on developing efficient theoretical frameworks and numerical methods. Among these, the monotone iterative technique has emerged as an effective approach, with demonstrated success across diverse application domains.

At its core, the monotone iterative technique establishes well-ordered iterative sequences that, when integrated with upper–lower solution theory, not only prove the existence of solutions but also provide a convergent pathway for numerical approximation. This methodology has been applied across various contexts: for instance, initial value problems involving Riemann–Liouville fractional differential equations [17], initial value problems and boundary value problems for Caputo fractional differential equations [15,18,19], initial value problems for systems involving Riemann–Liouville fractional derivatives [20], integral boundary value problems for Riemann–Liouville fractional differential equations [21], and initial value problems and boundary value problems for ordinary differential equations [22]. In [15], Ali et al. studied the following nonlinear boundary value problem (BVP) of FDEs of the form

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\phi \left(t\right)+f(t,\phi \left(t\right))=0,t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)=0,\phi \left(1\right)=0,{\phi }^{″}\left(0\right)=0,}\hfill \end{array}\right.$$

(2)

where $2<\nu \le 3$, ${}^{C}D_{0^{+}}^{\nu }$ is the Caputo fractional derivative and $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ is continuous. By using the method of the monotone iterative technique together with upper and lower solutions, the existence result of solutions is established.

However, the nonlinear term in [15] must satisfy the monotonicity conditions. In reality, nonlinear fractional differential equations with non-monotone terms offer a better representation of phenomena governed by objective laws. Therefore, relaxing these monotonicity constraints is crucial.

Motivated by the aforementioned discussions, this paper aims to further investigate the fractional order boundary value problem (2). In this study, we establish two novel comparison principles to circumvent the restrictive monotonicity requirements imposed on the nonlinear term. Subsequently, by employing the monotone iterative technique coupled with the method of upper and lower solutions, we prove the existence of extremal solutions to the problem (2).

2. Preliminaries

In this part, we first give the basic definitions, lemmas, and theorems.

Definition 1 

([11,12]). The integral with fractional order $\alpha >0$ of Riemann–Liouville type is defined for the function σ as

$$I_{0+}^{\alpha }\sigma \left(t\right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^t\sigma \left(s\right){(t-s)}^{\alpha -1}ds,$$

provided that the right-hand side is pointwise defined on $(0,+\infty )$, where $\Gamma \left(\alpha \right)$ is a Euler gamma function.

Definition 2 

([11,12]). The derivative with fractional order $\alpha >0$ of Riemann–Liouville type is defined for the function σ as

$$D_{0+}^{\alpha }\sigma \left(t\right)={\displaystyle \frac{1}{\Gamma (n-\alpha )}}{\left({\displaystyle \frac{d}{dt}}\right)}^{\left(n\right)}{\int }_0^t{\displaystyle \frac{\sigma \left(s\right)}{{(t-s)}^{\alpha +1-n}}}ds,n=\left[\alpha \right]+1,$$

provided that the right-hand side is pointwise defined on $(0,+\infty )$, where $\left[\alpha \right]$ denotes the integer part of the number α.

Definition 3 

([11,12]). For a function $f:[0,\infty )\to \mathbb{R}$, the Caputo derivative of fractional order α is defined as

$${}^{C}D_{0^{+}}^{\alpha }f\left(t\right)={\displaystyle \frac{1}{\Gamma (n-\alpha )}}{\int }_0^t{f}^{\left(n\right)}\left(s\right){(t-s)}^{n-\alpha -1}ds,n=\left[\alpha \right]+1,$$

where $\left[\alpha \right]$ denotes the integer part of the real number α.

Lemma 1 

([11,12]). Suppose that ${}^{C}D_{0^{+}}^{\alpha }u\in C(0,1)\cap L^1(0,1)$; then

$$I_{0+}^{\alpha }{}^{C}D_{0^{+}}^{\alpha }u\left(t\right)=u\left(t\right)+c_{1}t+c_2{t}^2+\dots +c_n{t}^n,$$

$c_j\in \mathbb{R}$, $j=1,2,\dots ,n$, where n is the smallest integer greater than or equal to α.

For brevity, let us take $E=\{u:u\in C[0,1]{,}^C{D}_{0^{+}}^{\nu }u\in C[0,1]\}$. In the Banach space $C[0,1]$, in which the norm is defined by $\parallel \phi \parallel ={max}_{t\in [0,1]}\left|\phi \left(t\right)\right|$, we set

$$P=\left\{\phi \in C[0,1]\mid \phi \left(t\right)\ge 0,\forall t\in [0,1]\right\}.$$

Thus, P is a positive cone in $C[0,1]$. Throughout this paper, the partial ordering is always given by P.

The following lemma establishes the existence and uniqueness of the solution for the linear boundary value problem, which is crucial for our later analysis.

Lemma 2. 

The linear boundary value problem

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\phi \left(t\right)+\alpha \left(t\right)=0,t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)=A_1,\phi \left(1\right)=A_2,{\phi }^{″}\left(0\right)=-A_3,}\hfill \end{array}\right.$$

(3)

where $A_1,A_2,A_3\in \mathbb{R}$ and $\alpha \in C[0,1]$, has the following integral representation of the solution:

$$\phi \left(t\right)={\int }_0^{1}G(t,s)\alpha \left(s\right)ds+A_1(1-t)+A_{2}t+{\displaystyle \frac{A_3}{2}}t(1-t),$$

where

$$G(t,s)=\left\{\begin{array}{cc}{\displaystyle \frac{t{(1-s)}^{\nu -1}-{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)},}\hfill & 0\le s\le t\le 1,\hfill \\ {\displaystyle \frac{t{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)},}\hfill & 0\le t\le s\le 1.\hfill \end{array}\right.$$

(4)

Proof. 

In view of Lemma 1, the linear BVP yields that

$$\begin{array}{cc}\hfill {\displaystyle \phi \left(t\right)}& {\displaystyle =-I_{0^{+}}^{\nu }\alpha \left(t\right)+\sum _{i=0}^{\left[\nu \right]}\frac{{\phi }^{\left(i\right)}\left(0\right)}{i!}{t}^i}\hfill \\ & {\displaystyle =-{\int }_0^t\frac{{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)}\alpha \left(s\right)ds+A_1+\phi \prime \left(0\right)t-\frac{A_3}{2}{t}^2,t\in (0,1).}\hfill \end{array}$$

On the other hand, $\phi \left(1\right)=A_2$ implies

$$A_2=-{\int }_0^1{\displaystyle \frac{{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)}}\alpha \left(s\right)ds+A_1+\phi \prime \left(0\right)-{\displaystyle \frac{A_3}{2}},$$

which after some manipulations yields

$$\phi \prime \left(0\right)=A_2+{\int }_0^1{\displaystyle \frac{{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)}}\alpha \left(s\right)ds-A_1+{\displaystyle \frac{A_3}{2}}.$$

Hence,

$$\begin{array}{cc}\hfill {\displaystyle \phi \left(t\right)}& {\displaystyle =-I_{0^{+}}^{\nu }\phi \left(t\right)+\sum _{i=0}^{\left[\nu \right]}\frac{{\phi }^{\left(i\right)}\left(0\right)}{i!}{t}^i=-{\int }_0^t\frac{{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)}\alpha \left(s\right)ds}\hfill \\ & {\displaystyle +A_1+\left(A_2+{\int }_0^1\frac{{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)}\alpha \left(s\right)ds-A_1+\frac{A_3}{2}\right)t-\frac{A_3}{2}{t}^2}\hfill \\ & {\displaystyle ={\int }_0^{1}G(t,s)\alpha \left(s\right)ds+A_1(1-t)+A_{2}t+\frac{A_3}{2}t(1-t).}\hfill \end{array}$$

This completes the proof. □

Lemma 3 

([14]). The Green’s function $G(t,s)$ defined by (4) satisfies

(i) 

$G(t,s)\ge 0$, for $t,s\in [0,1];$

(ii) 

$(t-t^{\nu -1})\psi \left(s\right)\le G(t,s)\le \psi \left(s\right)$, for $t,s\in [0,1]$, where

$$\psi \left(s\right)={\displaystyle \frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}}.$$

Lemma 4. 

Let G be the Green’s function defined by (4). Then

(i) 

$G(t,s)\le t\psi \left(s\right)$, for $t,s\in [0,1]$;

(ii) 

$G(t,s)\le {\displaystyle \frac{t(1-t){(1-s)}^{\nu -2}}{\Gamma (\nu -1)}},t,s\in [0,1];$

(iii) 

$\underset{t\in [0,1]}{max}{\int }_0^{1}G(t,s)ds={\displaystyle \frac{\nu -1}{{\nu }^{\frac{\nu }{\nu -1}}\Gamma (\nu +1)}};$

(iv) 

$G(t,s)\ge (\nu -2)t(1-t)\psi \left(s\right)$, $t,s\in [0,1].$

Proof. 

(i)

If $s\le t$,

$$G(t,s)={\displaystyle \frac{t{(1-s)}^{\nu -1}-{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)}}\le {\displaystyle \frac{t{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}}.$$

If $t\le s$,

$$G(t,s)={\displaystyle \frac{t{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}}.$$

(ii)

If $s\le t$,

$$\begin{array}{cc}\hfill {\displaystyle G(t,s)}& {\displaystyle =\frac{t{(1-s)}^{\nu -1}-{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)}\le \frac{t{(1-s)}^{\nu -1}-t{(t-s)}^{\nu -1}}{\Gamma \left(\nu \right)}}\hfill \\ \hfill & {\displaystyle =\frac{t(\nu -1){\int }_{t-s}^{1-s}{u}^{\nu -2}du}{\Gamma \left(\nu \right)}\le \frac{t(\nu -1){(1-s)}^{\nu -2}{\int }_{t-s}^{1-s}du}{\Gamma \left(\nu \right)}}\hfill \\ \hfill & {\displaystyle =\frac{(\nu -1)t(1-t){(1-s)}^{\nu -2}}{\Gamma \left(\nu \right)}=\frac{t(1-t){(1-s)}^{\nu -2}}{\Gamma (\nu -1)}.}\hfill \end{array}$$

If $t\le s$,

$$\begin{array}{cc}\hfill {\displaystyle G(t,s)}& ={\displaystyle \frac{t{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}}\le {\displaystyle \frac{t(1-t){(1-s)}^{\nu -2}}{\Gamma \left(\nu \right)}}\hfill \\ \hfill & {\displaystyle \le \frac{(\nu -1)t(1-t){(1-s)}^{\nu -2}}{\Gamma \left(\nu \right)}=\frac{t(1-t){(1-s)}^{\nu -2}}{\Gamma (\nu -1)}.}\hfill \end{array}$$

(iii)

Since

$${\int }_0^{1}G(t,s)ds={\displaystyle \frac{t-t^{\nu }}{\Gamma (\nu +1)}}:=\theta \left(t\right),t\in [0,1],$$

differentiating the function $\theta$ above, we immediately find that its maximum is achieved at the point

$$t^{*}={\displaystyle \frac{1}{{\nu }^{\frac{1}{\nu -1}}}}.$$

Moreover,

$$\theta \left(t^{*}\right)={\displaystyle \frac{\nu -1}{{\nu }^{\frac{\nu }{\nu -1}}\Gamma (\nu +1)}}.$$

Thus,

$$\underset{t\in [0,1]}{max}{\int }_0^{1}G(t,s)ds={\displaystyle \frac{\nu -1}{{\nu }^{\frac{\nu }{\nu -1}}\Gamma (\nu +1)}}.$$

(iv)

Define $g:[0,1]\to \mathbb{R}$ by

$$g\left(t\right)=\left\{\begin{array}{c}{\displaystyle \frac{1-t^{\nu -2}}{1-t},t\in [0,1),}\hfill \\ {\displaystyle \nu -2=\underset{t\to 1^{-}}{lim}g\left(t\right),t=1.}\hfill \end{array}\right.$$

Differentiating the function g, we immediately find that its critical point $t_0\in (0,1)$ (if it exists) satisfies

$${\displaystyle \frac{1-t_0^{\nu -2}}{1-t_0}}=(\nu -2)t_0^{\nu -3}.$$

Through the continuity of g on $[0,1],$ we obtain that

$$g\left(t\right)={\displaystyle \frac{1-t^{\nu -2}}{1-t}}\ge min\{g\left(0\right),g\left(t_0\right),g\left(1\right)\}=\nu -2,t\in [0,1).$$

Thus,

$$t-t^{\nu -1}\ge (\nu -2)t(1-t).,t\in [0,1)$$

This completes the proof. □

Lemma 5. 

Suppose that $\alpha \in C[0,1]$ and the constant M satisfies the following inequality:

$$0\le M<{\displaystyle \frac{{\nu }^{\frac{\nu }{\nu -1}}\Gamma (\nu +1)}{\nu -1}}.$$

(5)

Then the boundary value problem

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\phi \left(t\right)-M\phi \left(t\right)+\alpha \left(t\right)=0,t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)=A_1,\phi \left(1\right)=A_2,{\phi }^{″}\left(0\right)=-A_3}\hfill \end{array}\right.$$

(6)

has a unique continuous solution ${\phi }^{*}$, given by the expression

$${\phi }^{*}\left(t\right)=w_0\left(t\right)+{\int }_0^{1}Q(t,s)w_0\left(s\right)ds+{\int }_0^{1}H(t,s)\alpha \left(s\right)ds,$$

(7)

where

$$w_0\left(t\right)=A_1(1-t)+A_{2}t+{\displaystyle \frac{A_3}{2}}(t-t^2),$$

$$Q(t,s)=\sum _{i=1}^{\infty }{K}_i(t,s),K_1(t,s)=-MG(t,s),$$

$$\begin{array}{cc}& {\displaystyle K_i(t,s)}\hfill \\ \hfill =& {\displaystyle {(-M)}^i{\int }_0^1\dots {\int }_0^{1}G(t,s_1)G(s_1,s_2)\dots G(s_{i-1},s)d{s}_1\dots d{s}_{i-1},i=2,3,\dots ,}\hfill \end{array}$$

and

$$H(t,s)=G(t,s)+{\int }_0^{1}Q(t,\tau )G(\tau ,s)d\tau .$$

Proof. 

Using Lemma 2, it is easy to show that problem (6) is equivalent to the following integral equation:

$$\phi \left(t\right)={\int }_0^{1}G(t,s)(-M\phi \left(s\right)+\alpha \left(s\right))ds+A_1(1-t)+A_{2}t+{\displaystyle \frac{A_3}{2}}t(1-t),$$

i.e.,

$$\phi \left(t\right)=-M{\int }_0^{1}G(t,s)\phi \left(s\right)ds+{\phi }_0\left(t\right),$$

where

$${\phi }_0\left(t\right)={\omega }_0\left(t\right)+{\int }_0^{1}G(t,s)\alpha \left(s\right)ds,$$

$${\omega }_0\left(t\right)=A_1(1-t)+A_{2}t+{\displaystyle \frac{A_3}{2}}t(1-t).$$

Define the operator $T:C[0,1]\to C[0,1]$ by

$$\left(T\phi \right)\left(t\right)=-M{\int }_0^{1}G(t,s)\phi \left(s\right)ds+{\phi }_0\left(t\right).$$

Clearly, T is an operator from $C[0,1]$ into $C[0,1]$, and the fixed points of operator T coincide with the solutions of problem (6) by Lemma 2. We will indicate that the operator T has a unique fixed point. Let $\varphi ,\phi \in C[0,1]$; we have

$$\begin{array}{cc}\hfill {\displaystyle \parallel T\phi -T\varphi \parallel }& {\displaystyle =\underset{t\in [0,1]}{max}|\left(T\phi \right)\left(t\right)-\left(T\varphi \right)\left(t\right)|}\hfill \\ & {\displaystyle =\underset{t\in [0,1]}{max}\left|{\int }_0^{1}G(t,s)(M\phi \left(s\right)-M\varphi \left(s\right))ds\right|}\hfill \\ & {\displaystyle \le M\underset{t\in [0,1]}{max}{\int }_0^{1}G(t,s)ds\parallel \phi -\varphi \parallel }\hfill \\ & {\displaystyle \le \frac{M(\nu -1)}{{\nu }^{\frac{\nu }{\nu -1}}\Gamma (\nu +1)}\parallel \phi -\varphi \parallel .}\hfill \end{array}$$

The condition $0\le M<{\displaystyle \frac{{\nu }^{\frac{\nu }{\nu -1}}\Gamma (\nu +1)}{\nu -1}}$ implies that T is a contract operator on $C[0,1]$. Therefore, the operator T has a unique fixed point ${\phi }^{*}$ by the Banach fixed point theorem, which guarantees the existence and uniqueness of solution ${\phi }^{*}$ for (6), which can be obtained by calculating the sequence of successive approximations given by

$${\phi }_0\left(t\right)={\int }_0^{1}G(t,s)\alpha \left(s\right)ds+A_1(1-t)+A_{2}t+{\displaystyle \frac{A_3}{2}}t(1-t),$$

$$\begin{array}{cc}\hfill {\displaystyle {\phi }_n\left(t\right)}& {\displaystyle =\left(T{\phi }_{n-1}\right)\left(t\right)=-M{\int }_0^{1}G(t,s){\phi }_{n-1}\left(s\right)ds+{\phi }_0\left(t\right)}\hfill \\ \hfill & {\displaystyle ={\int }_0^1\sum _{i=1}^n{K}_i(t,s){\phi }_0\left(s\right)ds+{\phi }_0\left(t\right),t\in [0,1],n=1,2,\dots .}\hfill \end{array}$$

(8)

Note that the above sequence $\left\{{\phi }_n\right\}$ converges in the norm, hence uniformly on $[0,1]$ to the solution ${\phi }^{*}$. Taking the limit on both sides of (8), an explicit expression of the analytical solution ${\phi }^{*}$ of that is derived as follows:

$$\begin{array}{cc}\hfill {\displaystyle {\phi }^{*}\left(t\right)}& {\displaystyle ={\int }_0^{1}Q(t,s){\phi }_0\left(s\right)ds+{\phi }_0\left(t\right)}\hfill \\ & {\displaystyle ={\int }_0^{1}H(t,s)\alpha \left(s\right)ds+{\int }_0^{1}Q(t,s)w_0\left(s\right)ds+w_0\left(t\right),}\hfill \end{array}$$

where

$$Q(t,s)=\sum _{i=1}^{\infty }{K}_i(t,s),H(t,s)=G(t,s)+{\int }_0^{1}Q(t,\tau )G(\tau ,s)d\tau .$$

□

During the proof of Lemma 5, the solution to problem (6) was formulated as a fixed point of the contraction operator T. The existence and exact solution of problem (6) were then established via the Banach fixed point theorem and iterative methods, respectively. In fact, the solution to problem (6) can also be viewed as a solution to the following operator equation:

$$(I+MS)\phi ={\phi }_0,$$

where I is the identity operator, and S is the linear operator defined by

$$\left(S\phi \right)\left(t\right)={\int }_0^{1}G(t,s)\phi \left(s\right)ds,$$

and

$${\phi }_0\left(t\right)={\omega }_0\left(t\right)+{\int }_0^{1}G(t,s)\alpha \left(s\right)ds={\omega }_0\left(t\right)+\left(S\alpha \right)\left(t\right).$$

Condition (5) implies that the norm of the linear operator $MS$ satisfies $\parallel MS\parallel <1$. This guarantees the existence of the inverse operator ${(I+MS)}^{-1}$, which can be expressed as ${(I+MS)}^{-1}=\sum _{n=0}^{\infty }{(-MS)}^n$. Thus, the solution to problem (6) is

$$\begin{array}{cc}\hfill {\displaystyle \phi \left(t\right)}& {\displaystyle ={(I+MS)}^{-1}{\phi }_0\left(t\right)=\sum _{n=0}^{\infty }{(-MS)}^n{\phi }_0\left(t\right)}\hfill \\ \hfill =& {\displaystyle \sum _{n=0}^{\infty }{(-MS)}^{2n}{\phi }_0\left(t\right)+\sum _{n=0}^{\infty }{(-MS)}^{2n+1}{\phi }_0\left(t\right)}\hfill \\ \hfill =& {\displaystyle \sum _{n=0}^{\infty }{\left(MS\right)}^{2n}{\phi }_0\left(t\right)-\sum _{n=0}^{\infty }{\left(MS\right)}^{2n+1}{\phi }_0\left(t\right)}\hfill \\ \hfill =& {\displaystyle \sum _{n=0}^{\infty }{\left(MS\right)}^{2n}(I-MS){\phi }_0\left(t\right)}\hfill \\ \hfill =& {\displaystyle \sum _{n=0}^{\infty }{\left(MS\right)}^{2n}(I-MS)w_0\left(t\right)+\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}(I-MS)\left(S\alpha \right)\left(t\right).}\hfill \end{array}$$

(9)

Now, we establish several comparison principles.

Lemma 6 

(Comparison principle). Suppose that $\phi \in E$ satisfies

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\phi \left(t\right)-M\phi \left(t\right)\ge 0,t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)\le 0,\phi \left(1\right)\le 0,{\phi }^{″}\left(0\right)\ge 0,}\hfill \end{array}\right.$$

where M satisfies (5), as well as

$${\displaystyle \frac{M(\nu -1)}{\Gamma (\nu +1)}}-{\displaystyle \frac{K}{(\nu +1)\Gamma \left(\nu \right)}}\le 1,$$

(10)

$${\displaystyle \frac{M}{\Gamma (\nu +1)}}-{\displaystyle \frac{K}{\Gamma (\nu +2)}}\le 1,$$

(11)

$${\displaystyle \frac{M(\nu -1)}{\Gamma (\nu +2)}}-{\displaystyle \frac{K\nu }{\Gamma (\nu +3)}}\le 1.$$

(12)

Furthermore, one of the following conditions must hold: either (13) if $K\ge 0$, or (14) if $K<0$, where

$$\nu -{\displaystyle \frac{M}{\Gamma (\nu +1)}}+K\left({\displaystyle \frac{1}{\Gamma (\nu +2)}}-{\displaystyle \frac{\Gamma \left(\nu \right)}{\Gamma \left(2\nu \right)}}\right)\ge 2,$$

(13)

$$\nu -{\displaystyle \frac{M}{\Gamma (\nu +1)}}+{\displaystyle \frac{K}{\Gamma (\nu +2)}}\ge 2,$$

(14)

and

$$K={\displaystyle \frac{(\nu -2)M^2\left(\frac{1}{\Gamma (\nu +2)}-\frac{\Gamma \left(\nu \right)}{\Gamma \left(2\nu \right)}\right)}{1-{\left(\frac{M}{\Gamma (\nu +2)}-\frac{M\Gamma \left(\nu \right)}{\Gamma \left(2\nu \right)}\right)}^2}}-{\displaystyle \frac{M^3\frac{\nu -1}{\Gamma (\nu +2)}}{\Gamma (\nu +1)\left(1-{\left(\frac{M(\nu -1)}{\Gamma (\nu +2)}\right)}^2\right)}}.$$

(15)

Then, $\phi \left(t\right)\le 0$ for $t\in [0,1]$.

Proof. 

It is evident that there exist $\alpha \left(t\right)\ge 0$ and constants $A_1\le 0,A_2\le 0,A_3\le 0$ such that

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\phi \left(t\right)-M\phi \left(t\right)-\alpha \left(t\right)=0,t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)=A_1,\phi \left(1\right)=A_2,\phi \prime \prime \left(0\right)=-A_3.}\hfill \end{array}\right.$$

By Lemma 5, we have

$$\begin{array}{cc}\hfill {\displaystyle \phi \left(t\right)}& {\displaystyle =w_0\left(t\right)+{\int }_0^{1}Q(t,s)w_0\left(s\right)ds-{\int }_0^{1}H(t,s)\alpha \left(s\right)ds}\hfill \\ \hfill & {\displaystyle =A_1\left(w_1\left(t\right)+{\int }_0^{1}Q(t,s)w_1\left(s\right)ds\right)+A_2\left(w_2\left(t\right)+{\int }_0^{1}Q(t,s)w_2\left(s\right)ds\right)}\hfill \\ \hfill & {\displaystyle +\frac{A_3}{2}\left(w_3\left(t\right)+{\int }_0^{1}Q(t,s)w_3\left(s\right)ds\right)-{\int }_0^{1}H(t,s)\alpha \left(s\right)ds}\hfill \\ \hfill & {\displaystyle :=A_1{J}_1\left(t\right)+A_2{J}_2\left(t\right)+\frac{A_3}{2}{J}_3\left(t\right)-J_4\left(t\right),}\hfill \end{array}$$

where

$$w_0\left(t\right)=A_1{w}_1\left(t\right)+A_2{w}_2\left(t\right)+{\displaystyle \frac{A_3}{2}}{w}_3\left(t\right),$$

$$w_1\left(t\right)=(1-t),w_2\left(t\right)=t,w_3\left(t\right)=t(1-t).$$

Since $\alpha \left(t\right)\ge 0$ and $A_1,A_2,A_3\le 0$, it suffices to establish $\phi \left(t\right)\le 0$ by verifying $J_i\left(t\right)\ge 0$ for $i=1,2,3$ and $H(t,s)\ge 0$ for $t,s\in [0,1]$, where $H(t,s)$ defines the term $J_4\left(t\right)$. To this end, we first investigate the properties of $Q(t,s)$. For $i\ge 2$, by applying Lemma 4 (ii) to the first $i-1$ factors in the integral below, and Lemma 4 (i) to the final factor, we obtain

$$\begin{array}{cc}& {\displaystyle {(-1)}^i{K}_i(t,s)}\hfill \\ \hfill =& {\displaystyle M^i{\int }_0^1\dots {\int }_0^{1}G(t,s_1)G(s_1,s_2)\dots G(s_{i-2},s_{i-1})G(s_{i-1},s)d{s}_1\dots d{s}_{i-1}}\hfill \\ \hfill \le & {\displaystyle M^i{\int }_0^1\dots {\int }_0^1\frac{t(1-t){(1-s_1)}^{\nu -2}}{\Gamma (\nu -1)}·\frac{s_1(1-s_1){(1-s_2)}^{\nu -2}}{\Gamma (\nu -1)}}\hfill \\ & {\displaystyle \dots \frac{s_{i-2}(1-s_{i-2}){(1-s_{i-1})}^{\nu -2}}{\Gamma (\nu -1)}·\frac{s_{i-1}{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}d{s}_1\dots d{s}_{i-1}}\hfill \\ \hfill =& {\displaystyle M^i\frac{t(1-t){(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}{\int }_0^1\frac{s_1(1-s_1){(1-s_1)}^{\nu -2}}{\Gamma (\nu -1)}d{s}_1·{\int }_0^1\frac{s_2(1-s_2){(1-s_2)}^{\nu -2}}{\Gamma (\nu -1)}d{s}_2}\hfill \\ & {\displaystyle \dots {\int }_0^1\frac{s_{i-2}(1-s_{i-2}){(1-s_{i-2})}^{\nu -2}}{\Gamma (\nu -1)}d{s}_{i-2}·{\int }_0^1\frac{s_{i-1}{(1-s_{i-1})}^{\nu -2}}{\Gamma (\nu -1)}d{s}_{i-1}}\hfill \\ \hfill =& {\displaystyle M^i\frac{t(1-t)}{\Gamma (\nu +1)}\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}{\left[\frac{\nu -1}{\Gamma (\nu +2)}\right]}^{i-2}.}\hfill \end{array}$$

For $i\ge 2$, by applying Lemma 4 (iv) to the first factor in the integral below, and Lemma 3 (ii) to the remaining $i-1$ factors, we obtain

$$\begin{array}{cc}& {\displaystyle {(-1)}^i{K}_i(t,s)}\hfill \\ \hfill =& {\displaystyle M^i{\int }_0^1\dots {\int }_0^{1}G(t,s_1)G(s_1,s_2)\dots G(s_{i-1},s)d{s}_1\dots d{s}_{i-1}}\hfill \\ \hfill \ge & {\displaystyle M^i{\int }_0^1\dots {\int }_0^1(\nu -2)t(1-t)\frac{{(1-s_1)}^{\nu -1}}{\Gamma \left(\nu \right)}(s_1-s_1^{\nu -1})·\frac{{(1-s_2)}^{\nu -1}}{\Gamma \left(\nu \right)}}\hfill \\ & {\displaystyle \dots (s_{i-1}-s_{i-1}^{\nu -1})\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}d{s}_1\dots d{s}_{i-1}}\hfill \\ \hfill =& {\displaystyle M^i(\nu -2)\frac{t(1-t){(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}{\left[{\int }_0^1\frac{(s_1-s_1^{\nu -1}){(1-s_1)}^{\nu -1}}{\Gamma \left(\nu \right)}d{s}_1\right]}^{i-1}}\hfill \\ \hfill =& {\displaystyle M^i(\nu -2)\frac{t(1-t){(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}{\left[\frac{1}{\Gamma (\nu +2)}-\frac{\Gamma \left(\nu \right)}{\Gamma \left(2\nu \right)}\right]}^{i-1}.}\hfill \end{array}$$

Combining the above two inequalities with Lemma 4 (i), we obtain the following properties of the function $Q(t,s)$:

$$\begin{array}{cc}& {\displaystyle Q(t,s)=\sum _{i=1}^{\infty }{K}_i(t,s)=K_1(t,s)+\sum _{i=1}^{\infty }{K}_{2i}(t,s)+\sum _{i=1}^{\infty }{K}_{2i+1}(t,s)}\hfill \\ \hfill =& {\displaystyle -MG(t,s)+\sum _{i=1}^{\infty }{(-1)}^{2i}{K}_{2i}(t,s)-\sum _{i=1}^{\infty }{(-1)}^{2i+1}{K}_{2i+1}(t,s)}\hfill \\ \hfill \ge & {\displaystyle -M\frac{t(1-t){(1-s)}^{\nu -2}}{\Gamma (\nu -1)}+\sum _{i=1}^{\infty }{M}^{2i}(\nu -2)\frac{t(1-t){(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}{\left[\frac{1}{\Gamma (\nu +2)}-\frac{\Gamma \left(\nu \right)}{\Gamma \left(2\nu \right)}\right]}^{2i-1}}\hfill \\ & {\displaystyle -\sum _{i=1}^{\infty }{M}^{2i+1}\frac{t(1-t)}{\Gamma (\nu +1)}\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}{\left[\frac{\nu -1}{\Gamma (\nu +2)}\right]}^{2i-1}}\hfill \\ \hfill =& {\displaystyle -M\frac{t(1-t){(1-s)}^{\nu -2}}{\Gamma (\nu -1)}+\frac{K}{\Gamma \left(\nu \right)}t(1-t){(1-s)}^{\nu -1}.}\hfill \end{array}$$

Based on the above inequality, we investigate the non-negativity of $J_i\left(t\right)$ (i = 1, 2, 3, 4), respectively. For $J_1\left(t\right)$, by (10), we have

$$\begin{array}{cc}\hfill {\displaystyle J_1\left(t\right)=}& {\displaystyle w_1\left(t\right)+{\int }_0^{1}Q(t,s)w_1\left(s\right)ds}\hfill \\ \hfill \ge & {\displaystyle (1-t)+{\int }_0^1\left(-M\frac{t(1-t){(1-s)}^{\nu -2}}{\Gamma (\nu -1)}+\frac{K}{\Gamma \left(\nu \right)}t(1-t){(1-s)}^{\nu -1}\right)(1-s)ds}\hfill \\ \hfill =& {\displaystyle (1-t)-M\frac{t(1-t)}{\Gamma (\nu -1)}{\int }_0^1{(1-s)}^{\nu -2}(1-s)ds}\hfill \\ & {\displaystyle +\frac{K}{\Gamma \left(\nu \right)}t(1-t){\int }_0^1{(1-s)}^{\nu -1}(1-s)ds}\hfill \\ \hfill =& {\displaystyle (1-t)-M\frac{t(1-t)}{\nu \Gamma (\nu -1)}+\frac{K}{(\nu +1)\Gamma \left(\nu \right)}t(1-t)}\hfill \\ \hfill \ge & {\displaystyle t(1-t)-M\frac{t(1-t)}{\nu \Gamma (\nu -1)}+\frac{K}{(\nu +1)\Gamma \left(\nu \right)}t(1-t)}\hfill \\ \hfill =& {\displaystyle t(1-t)\left[1-\frac{M}{\nu \Gamma (\nu -1)}+\frac{K}{(\nu +1)\Gamma \left(\nu \right)}\right]}\hfill \\ \hfill =& {\displaystyle t(1-t)\left[1-\frac{M(\nu -1)}{\Gamma (\nu +1)}+\frac{K}{(\nu +1)\Gamma \left(\nu \right)}\right]\ge 0.}\hfill \end{array}$$

For $J_2\left(t\right)$, by (11), we have

$$\begin{array}{cc}\hfill {\displaystyle J_2\left(t\right)=}& {\displaystyle w_2\left(t\right)+{\int }_0^{1}Q(t,s)w_2\left(s\right)ds}\hfill \\ \hfill \ge & {\displaystyle t+{\int }_0^1\left(-M\frac{t(1-t){(1-s)}^{\nu -2}}{\Gamma (\nu -1)}+\frac{K}{\Gamma \left(\nu \right)}t(1-t){(1-s)}^{\nu -1}\right)sds}\hfill \\ \hfill =& {\displaystyle t-M\frac{t(1-t)}{\Gamma (\nu -1)}{\int }_0^1{(1-s)}^{\nu -2}sds+\frac{K}{\Gamma \left(\nu \right)}t(1-t){\int }_0^1{(1-s)}^{\nu -1}sds}\hfill \\ \hfill =& {\displaystyle t-M\frac{t(1-t)}{\Gamma (\nu +1)}+\frac{K}{\Gamma (\nu +2)}t(1-t)}\hfill \\ \hfill \ge & {\displaystyle t(1-t)-M\frac{t(1-t)}{\Gamma (\nu +1)}+\frac{K}{\Gamma (\nu +2)}t(1-t)}\hfill \\ \hfill =& {\displaystyle t(1-t)\left[1-\frac{M}{\Gamma (\nu +1)}+\frac{K}{\Gamma (\nu +2)}\right]\ge 0.}\hfill \end{array}$$

For $J_3\left(t\right)$, by (12), we have

$$\begin{array}{cc}\hfill {\displaystyle J_3\left(t\right)=}& {\displaystyle w_3\left(t\right)+{\int }_0^{1}Q(t,s)w_3\left(s\right)ds}\hfill \\ \hfill \ge & {\displaystyle t(1-t)+{\int }_0^1\left(-M\frac{t(1-t){(1-s)}^{\nu -2}}{\Gamma (\nu -1)}+\frac{K}{\Gamma \left(\nu \right)}t(1-t){(1-s)}^{\nu -1}\right)s(1-s)ds}\hfill \\ \hfill =& {\displaystyle t(1-t)-M\frac{t(1-t)}{\Gamma (\nu -1)}{\int }_0^1{(1-s)}^{\nu -1}sds+\frac{K}{\Gamma \left(\nu \right)}t(1-t){\int }_0^1{(1-s)}^{\nu }sds}\hfill \\ \hfill =& {\displaystyle t(1-t)-M\frac{(\nu -1)t(1-t)}{\Gamma (\nu +2)}+\frac{K\nu }{\Gamma (\nu +3)}t(1-t)}\hfill \\ \hfill =& {\displaystyle t(1-t)\left[1-\frac{M(\nu -1)}{\Gamma (\nu +2)}+\frac{K\nu }{\Gamma (\nu +3)}\right]\ge 0.}\hfill \end{array}$$

We now proceed to prove the non-negativity of $H(t,s)$, which is divided into two cases depending on the sign of K.

Case 1: $K\ge 0$. In this scenario, it follows from (13) that

$$\begin{array}{cc}& {\displaystyle H(t,s)=G(t,s)+{\int }_0^{1}Q(t,\tau )G(\tau ,s)d\tau }\hfill \\ \hfill \ge & {\displaystyle G(t,s)+{\int }_0^1\left(-M\frac{t(1-t){(1-\tau )}^{\nu -2}}{\Gamma (\nu -1)}+\frac{K}{\Gamma \left(\nu \right)}t(1-t){(1-\tau )}^{\nu -1}\right)G(\tau ,s)d\tau }\hfill \\ \hfill =& {\displaystyle G(t,s)-Mt(1-t){\int }_0^1\frac{{(1-\tau )}^{\nu -2}}{\Gamma (\nu -1)}G(\tau ,s)d\tau +\frac{K}{\Gamma \left(\nu \right)}t(1-t){\int }_0^1{(1-\tau )}^{\nu -1}G(\tau ,s)d\tau }\hfill \\ \hfill \ge & {\displaystyle (\nu -2)t(1-t)\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}-Mt(1-t){\int }_0^1\frac{{(1-\tau )}^{\nu -2}}{\Gamma (\nu -1)}\frac{\tau {(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}d\tau }\hfill \\ & {\displaystyle +\frac{K}{\Gamma \left(\nu \right)}t(1-t){\int }_0^1{(1-\tau )}^{\nu -1}(\tau -{\tau }^{\nu -1})\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}d\tau }\hfill \\ \hfill =& {\displaystyle (\nu -2)t(1-t)\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}-M\frac{t(1-t)}{\Gamma (\nu +1)}\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}}\hfill \\ & {\displaystyle +Kt(1-t)\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}\left(\frac{1}{\Gamma (\nu +2)}-\frac{\Gamma \left(\nu \right)}{\Gamma \left(2\nu \right)}\right)}\hfill \\ \hfill =& {\displaystyle t(1-t)\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}\left((\nu -2)-\frac{M}{\Gamma (\nu +1)}+K\left(\frac{1}{\Gamma (\nu +2)}-\frac{\Gamma \left(\nu \right)}{\Gamma \left(2\nu \right)}\right)\right)\ge 0,}\hfill \end{array}$$

where the second inequality is obtained by applying Lemma 4 (iv) to the first and third terms, and Lemma 4 (i) to the second term. The final non-negativity directly follows from condition (13).

Case 2: $K<0$. By utilizing (14), the function $H(t,s)$ satisfies

$$\begin{array}{cc}& {\displaystyle H(t,s)=G(t,s)+{\int }_0^{1}Q(t,\tau )G(\tau ,s)d\tau }\hfill \\ \hfill \ge & {\displaystyle G(t,s)+{\int }_0^1\left(-M\frac{t(1-t){(1-\tau )}^{\nu -2}}{\Gamma (\nu -1)}+\frac{K}{\Gamma \left(\nu \right)}t(1-t){(1-\tau )}^{\nu -1}\right)G(\tau ,s)d\tau }\hfill \\ \hfill =& {\displaystyle G(t,s)-Mt(1-t){\int }_0^1\frac{{(1-\tau )}^{\nu -2}}{\Gamma (\nu -1)}G(\tau ,s)d\tau +\frac{K}{\Gamma \left(\nu \right)}t(1-t){\int }_0^1{(1-\tau )}^{\nu -1}G(\tau ,s)d\tau }\hfill \\ \hfill \ge & {\displaystyle (\nu -2)t(1-t)\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}-Mt(1-t){\int }_0^1\frac{{(1-\tau )}^{\nu -2}}{\Gamma (\nu -1)}\frac{\tau {(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}d\tau }\hfill \\ & {\displaystyle +\frac{K}{\Gamma \left(\nu \right)}t(1-t){\int }_0^1{(1-\tau )}^{\nu -1}\frac{\tau {(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}d\tau }\hfill \\ \hfill =& {\displaystyle (\nu -2)t(1-t)\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}-M\frac{t(1-t)}{\Gamma (\nu +1)}\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}+Kt(1-t)\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}\frac{1}{\Gamma (\nu +2)}}\hfill \\ \hfill =& {\displaystyle t(1-t)\frac{{(1-s)}^{\nu -1}}{\Gamma \left(\nu \right)}\left((\nu -2)-\frac{M}{\Gamma (\nu +1)}+\frac{K}{\Gamma (\nu +2)}\right)\ge 0,}\hfill \end{array}$$

where the second inequality is obtained by applying Lemma 4 (iv) to the first term, and Lemma 4 (i) to the latter two terms. The final non-negativity directly follows from condition (14).

Consequently, the above estimates yield $J_i\left(t\right)\ge 0$ for $i=1,2,3$ and $H(t,s)\ge 0$ for $t,s\in [0,1]$. Since $\alpha \left(t\right)\ge 0$, it is straightforward to see that $J_4\left(t\right)={\int }_0^{1}H(t,s)\alpha \left(s\right)ds\ge 0$. Recalling that $A_1,A_2,A_3\le 0$ and combining these non-negativities with $J_i\left(t\right)$, we immediately obtain $\phi \left(t\right)=A_1{J}_1\left(t\right)+A_2{J}_2\left(t\right)+{\displaystyle \frac{A_3}{2}}{J}_3\left(t\right)-J_4\left(t\right)\le 0$. This completes the proof. □

Since the parameter M in Lemma 6 is required to satisfy multiple conditions simultaneously, a natural question arises: does there exist an $M>0$ that fulfills all these constraints? In the following, we demonstrate the existence of a non-trivial interval for M.

First, based on Lemma 5, we establish a necessary upper bound for M:

$$M\in (0,N),N={\displaystyle \frac{{\nu }^{\frac{\nu }{\nu -1}}\Gamma (\nu +1)}{\nu -1}}.$$

Next, we analyze the asymptotic behavior of the constant K as $M\to 0^{+}$. Noting that

$$\underset{M\to 0^{+}}{lim}{\displaystyle \frac{K}{M^2}}=(\nu -2)\left({\displaystyle \frac{1}{\Gamma (\nu +2)}}-{\displaystyle \frac{\Gamma \left(\nu \right)}{\Gamma \left(2\nu \right)}}\right)\triangleq A>0,$$

it follows that there exists a constant $N_1\le N$ such that for all $M\in (0,N_1)$, we have $K>0$ and $K\sim A{M}^2$ (as $M\to 0^{+}$).

Finally, utilizing the asymptotic estimate $K\sim A{M}^2$, we evaluate the limits of the remaining expressions involved in Lemma 6 as $M\to 0^{+}$:

$$\begin{array}{cc}\hfill \underset{M\to 0^{+}}{lim}\left({\displaystyle \frac{M(\nu -1)}{\Gamma (\nu +1)}}-{\displaystyle \frac{K}{(\nu +1)\Gamma \left(\nu \right)}}\right)& =0<1,\hfill \\ \hfill \underset{M\to 0^{+}}{lim}\left({\displaystyle \frac{M}{\Gamma (\nu +1)}}-{\displaystyle \frac{K}{\Gamma (\nu +2)}}\right)& =0<1,\hfill \\ \hfill \underset{M\to 0^{+}}{lim}\left({\displaystyle \frac{M(\nu -1)}{\Gamma (\nu +2)}}-{\displaystyle \frac{K\nu }{\Gamma (\nu +3)}}\right)& =0<1,\hfill \\ \hfill \underset{M\to 0^{+}}{lim}\left(\nu -{\displaystyle \frac{M}{\Gamma (\nu +1)}}+K\left({\displaystyle \frac{1}{\Gamma (\nu +2)}}-{\displaystyle \frac{\Gamma \left(\nu \right)}{\Gamma \left(2\nu \right)}}\right)\right)& =\nu >2.\hfill \end{array}$$

Given that $\nu \in (2,3]$, all the above limits strictly satisfy the required inequalities in Lemma 6. By the local sign-preserving property of limits, there must exist a constant $N_2\in (0,N_1)$ such that for all $M\in (0,N_2)$, the constant K remains positive, and all the inequalities in Lemma 6 hold simultaneously.

Lemma 7. 

Suppose that $\phi \in E$ satisfies

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\phi \left(t\right)-M\phi \left(t\right)\ge 0,t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)\le 0,\phi \left(1\right)\le 0,\phi \prime \prime \left(0\right)\ge 0,}\hfill \end{array}\right.$$

where M satisfies (5) and

$$0\le M\le min\left\{{\displaystyle \frac{\Gamma (\nu +1)}{\nu -1}},(\nu -2)\Gamma (\nu +1)\right\}.$$

(16)

Then, $\phi \left(t\right)\le 0$ for $t\in [0,1]$.

Proof. 

Clearly, there exist $\alpha \left(t\right)\ge 0$ and constants $A_1\le 0,A_2\le 0,A_3\le 0$ such that

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\phi \left(t\right)-M\phi \left(t\right)-\alpha \left(t\right)=0,t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)=A_1,\phi \left(1\right)=A_2,\phi \prime \prime \left(0\right)=-A_3.}\hfill \end{array}\right.$$

Based on the operator representation (9), $\phi \left(t\right)$ can be expressed as

$$\begin{array}{cc}\hfill {\displaystyle \phi \left(t\right)}& {\displaystyle =\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}(I-MS)w_0\left(t\right)-\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}(I-MS)\left(S\alpha \right)\left(t\right)}\hfill \\ \hfill & {\displaystyle =A_1\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}(I-MS)w_1\left(t\right)+A_2\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}(I-MS)w_2\left(t\right)}\hfill \\ \hfill & {\displaystyle +\frac{A_3}{2}\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}(I-MS)w_3\left(t\right)-\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}(I-MS)\left(S\alpha \right)\left(t\right)}\hfill \\ \hfill & {\displaystyle :=A_1\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}{I}_1\left(t\right)+A_2\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}{I}_2\left(t\right)}\hfill \\ \hfill & {\displaystyle +\frac{A_3}{2}\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}{I}_3\left(t\right)-\sum _{n=0}^{\infty }{\left(MS\right)}^{2n}{I}_4\left(t\right),}\hfill \end{array}$$

where $w_0\left(t\right)=A_1{w}_1\left(t\right)+A_2{w}_2\left(t\right)+{\displaystyle \frac{A_3}{2}}{w}_3\left(t\right)$ with $w_1\left(t\right)=(1-t)$, $w_2\left(t\right)=t$ and $w_3\left(t\right)=t(1-t)$.

Since the operator S is increasing, together with the facts that $\alpha \left(t\right)\ge 0$ and $A_1,A_2,A_3\le 0$, it suffices to establish $\phi \left(t\right)\le 0$ by verifying $I_i\left(t\right)\ge 0$ for $i=1,2,3,4$.

First, note that $\nu \in (2,3]$. By the properties of the Gamma function, we have

$${\displaystyle \frac{\nu -1}{\Gamma (\nu +1)}}>{\displaystyle \frac{1}{\Gamma (\nu +1)}},{\displaystyle \frac{\nu -1}{\Gamma (\nu +1)}}>{\displaystyle \frac{\nu -1}{\Gamma (\nu +2)}}.$$

Therefore, if a positive constant M satisfies ${\displaystyle \frac{M(\nu -1)}{\Gamma (\nu +1)}}\le 1$ it necessarily implies that

$${\displaystyle \frac{M}{\Gamma (\nu +1)}}\le 1,{\displaystyle \frac{M(\nu -1)}{\Gamma (\nu +2)}}\le 1.$$

(17)

We now establish the non-negativity of $I_i\left(t\right)$ (i=1,2,3,4) using the inequalities above. For $I_1\left(t\right)$, by Lemma 4 (ii) and (16), we have

$$\begin{array}{cc}\hfill I_1\left(t\right)& =(I-MS)w_1\left(t\right)=w_1\left(t\right)-{\int }_0^{1}G(t,s)w_1\left(s\right)ds\hfill \\ \hfill & =(1-t)-M{\int }_0^{1}G(t,s)(1-s)ds\hfill \\ \hfill & \ge t(1-t)-{\displaystyle \frac{Mt(1-t)}{\Gamma (\nu -1)}}{\int }_0^1{(1-s)}^{\nu -2}(1-s)ds\hfill \\ \hfill & =t(1-t)\left[1-{\displaystyle \frac{M}{\nu \Gamma (\nu -1)}}\right]=t(1-t)\left[1-{\displaystyle \frac{M(\nu -1)}{\Gamma (\nu +1)}}\right]\ge 0\hfill \end{array}$$

For $I_2\left(t\right)$, by Lemma 4 (ii) and (17), we have

$$\begin{array}{cc}\hfill I_2\left(t\right)& =(I-MS)w_2\left(t\right)=w_2\left(t\right)-M{\int }_0^{1}G(t,s)w_2\left(s\right)ds\hfill \\ \hfill & =t-M{\int }_0^{1}G(t,s)sds\ge t(1-t)-{\displaystyle \frac{Mt(1-t)}{\Gamma (\nu -1)}}{\int }_0^1{(1-s)}^{\nu -2}sds\hfill \\ \hfill & =t(1-t)\left[1-{\displaystyle \frac{M}{\Gamma (\nu +1)}}\right]\ge 0\hfill \end{array}$$

For $I_3\left(t\right)$, by Lemma 4 (ii) and (17), we have

$$\begin{array}{cc}\hfill I_3\left(t\right)& =(I-MS)w_3\left(t\right)=w_3\left(t\right)-M{\int }_0^{1}G(t,s)w_3\left(s\right)ds\hfill \\ \hfill & =t(1-t)-M{\int }_0^{1}G(t,s)s(1-s)ds\hfill \\ \hfill & \ge t(1-t)-{\displaystyle \frac{Mt(1-t)}{\Gamma (\nu -1)}}{\int }_0^1{(1-s)}^{\nu -2}s(1-s)ds\hfill \\ \hfill & =t(1-t)\left[1-{\displaystyle \frac{M(\nu -1)}{\Gamma (\nu +2)}}\right]\ge 0\hfill \end{array}$$

Recalling that $\alpha \left(t\right)\ge 0$, for $I_4\left(t\right)$, applying Lemma 4 (i), (ii) and (17) yields

$$\begin{array}{cc}\hfill I_4\left(t\right)& =(I-MS)\left(S\alpha \right)\left(t\right)=\left(S\alpha \right)\left(t\right)-M\left(S^2\alpha \right)\left(t\right)\hfill \\ \hfill & ={\int }_0^{1}G(t,s)\alpha \left(s\right)ds-M{\int }_0^1{\int }_0^{1}G(t,\tau )G(\tau ,s)\alpha \left(s\right)d\tau ds\hfill \\ \hfill & \ge (\nu -2)t(1-t){\int }_0^1\psi \left(s\right)\alpha \left(s\right)ds-{\displaystyle \frac{Mt(1-t)}{\Gamma (\nu -1)}}{\int }_0^1{\int }_0^1{(1-\tau )}^{\nu -2}\tau \psi \left(s\right)\alpha \left(s\right)d\tau ds\hfill \\ \hfill & =t(1-t){\int }_0^1\psi \left(s\right)\alpha \left(s\right)ds\left[(\nu -2)-{\displaystyle \frac{M}{\Gamma (\nu +1)}}\right]\ge 0.\hfill \end{array}$$

Consequently, the above estimates demonstrate that $I_i\left(t\right)\ge 0$ for $i=1,2,3,4$. Recalling that $A_1,A_2,A_3\le 0$ and combining these non-negativities with $I_i\left(t\right)$ terms, we immediately obtain $\phi \left(t\right)=A_1{I}_1\left(t\right)+A_2{I}_2\left(t\right)+{\displaystyle \frac{A_3}{2}}{I}_3\left(t\right)-I_4\left(t\right)\le 0$. This completes the proof. □

Between the two aforementioned comparison principles, the conditions imposed on the constant M in Lemma 7 are more straightforward to verify than those in Lemma 6. Moreover, extensive numerical evidence (though not yet rigorously proven theoretically) indicates that Lemma 7 yields a broader applicable range for M. Nevertheless, we retain Lemma 6 in this paper to explore a compelling prospect: whether a deeper analysis of the Green’s function’s properties could elevate the admissible range of M in Lemma 6 to match, or even surpass, that of Lemma 7.

When $\nu =3$, problem (2) reduces to the integer-order boundary value problem:

$$\left\{\begin{array}{c}{\displaystyle \phi \prime \prime \prime \left(t\right)+f(t,\phi \left(t\right))=0,t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)=0,\phi \left(1\right)=0,\phi ″\left(0\right)=0.}\hfill \end{array}\right.$$

(18)

Moreover, various results concerning the existence of positive solutions to (18) can be found in [23,24,25]. In this case, the comparison principle established in Lemma 7 restricts the parameter M to the interval $[0,3]$. Compared to integer-order operators, the properties of fractional order operators at extreme points are relatively underexplored; thus, the proof of Lemmas 6 and 7 heavily relies on operator equations and the representation of inverse operators. In fact, when $\nu =3$, we can alternatively leverage the standard properties of integer-order derivatives. To the best of our knowledge, a comparison principle tailored specifically to this integer-order scenario has not been previously reported in the literature. However, a significant drawback of this classical differential approach is that it yields a narrower admissible range, restricting M to the interval $[0,2]$ rather than $[0,3]$. Despite this limitation, we continue to state the comparison principle obtained through this classical differential approach below. The rationale for including it is that, initially, we did not fully grasp the specific drawbacks of this method, nor did we possess a clear strategy on how to improve it to achieve a wider range for M.

Lemma 8. 

Suppose that φ satisfies

$$\left\{\begin{array}{c}{\displaystyle \phi \prime \prime \prime \left(t\right)\ge M\phi \left(t\right),t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)\le 0,\phi \left(1\right)\le 0,\phi \prime \prime \left(0\right)\ge 0,}\hfill \end{array}\right.$$

where $0\le M\le 2$. Then, for $t\in [0,1]$, we have $\phi \left(t\right)\le 0$.

Proof. 

(1)

If $M=0$, then we have $\phi \prime \prime \prime \left(t\right)\ge 0$. This together with $\phi \prime \prime \left(0\right)\ge 0$ implies that $\phi \prime \prime \left(t\right)\ge 0$ for $t\in [0,1]$, which indicates that $\phi \left(t\right)$ is a convex function on $[0,1]$. Hence, we have

$$\phi \left(t\right)=\phi \left(\right(1-t)·0+t·1)\le (1-t)\phi \left(0\right)+t\phi \left(1\right)\le 0.$$

(2)

If $0<M\le 2$. Suppose there is $t_1\in (0,1)$ such that $\phi \left(t_1\right)=\underset{t\in [0,1]}{max}\phi \left(t\right)>0$, then $\phi \prime \left(t_1\right)=0$.

Let $m=\phi \left(t_2\right)=\underset{t\in [0,t_1]}{min}\phi \left(t\right)$. Then, we infer that $m<0$. In fact, if $m\ge 0$, by the continuity of $\phi$, there exist $\eta \in (t_1,1]$ such that

$$\phi \left(t\right)\ge 0,t\in [0,\eta ]and\phi \left(\eta \right)=0.$$

Then, by the non-negativity of the function and the definition of the derivative, we obtain $\phi \prime \left(0\right)\ge 0$. Note that $\phi \prime \prime \prime \left(t\right)\ge M\phi \left(t\right)\ge 0$ for $t\in [0,\eta ]$ and $\phi \prime \prime \left(0\right)\ge 0$; we get $\phi \prime \left(t\right)\ge 0$ for $t\in [0,\eta ]$. Thus, $\phi \left(t\right)$ is a nondecreasing function on $[0,\eta ]$. Consequently, $\phi \left(\eta \right)\ge \phi \left(t_1\right)>0$, which contradicts the fact $\phi \left(\eta \right)=0$.

According to Taylor’s formula, there exists a $t_3\in (t_2,t_1)$ such that

$$\phi \left(t_2\right)=\phi \left(t_1\right)+\phi \prime \left(t_1\right)(t_2-t_1)+{\displaystyle \frac{\phi \prime \prime \left(t_3\right)}{2}}{(t_2-t_1)}^2=\phi \left(t_1\right)+{\displaystyle \frac{\phi \prime \prime \left(t_3\right)}{2}}{(t_2-t_1)}^2,$$

and so that

$$\phi \prime \prime \left(t_3\right)={\displaystyle \frac{2(\phi \left(t_2\right)-\phi \left(t_1\right))}{{(t_2-t_1)}^2}}.$$

Using the mean value theorem again, there exist $t_4\in (0,t_3)$ such that

$$\phi \prime \prime \prime \left(t_4\right)={\displaystyle \frac{\phi \prime \prime \left(t_3\right)-\phi \prime \prime \left(0\right)}{t_3-0}}\le {\displaystyle \frac{\phi \prime \prime \left(t_4\right)}{t_3-0}}={\displaystyle \frac{2(\phi \left(t_2\right)-\phi \left(t_1\right))}{t_3{(t_2-t_1)}^2}}.$$

Since $m=\phi \left(t_2\right)=\underset{t\in [0,t_1]}{min}\phi \left(t\right)<0$ and $\phi \left(t_1\right)>0$, we obtain that

$$Mm\le M\phi \left(t_4\right)\le \phi \prime \prime \prime \left(t_4\right)={\displaystyle \frac{2(\phi \left(t_2\right)-\phi \left(t_1\right))}{t_3{(t_2-t_1)}^2}}<2m.$$

which implies $M>2$. This is a contradiction. Hence, $\phi \left(t\right)\le 0$ for $t\in [0,1]$ and the proof of Lemma is complete. □

3. Main Results

In this section, we will prove the existence of extremal solutions of (2) by using Lemmas 5 and 7.

First, we list the definitions of upper and lower solutions below for problem (2).

Definition 4. 

A function $u_0\in E$ is called a lower solution of the problem (2) if it satisfies

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }{u}_0\left(t\right)+f(t,u_0\left(t\right))\ge 0,t\in (0,1),}\hfill \\ {\displaystyle u_0\left(0\right)\le 0,u_0\left(1\right)\le 0,u_0\prime \prime \left(0\right)\ge 0.}\hfill \end{array}\right.$$

Analogously, the function $v_0\in E$ is called an upper solution of problem (2) if it satisfies

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }{v}_0\left(t\right)+f(t,v_0\left(t\right))\le 0,t\in (0,1),}\hfill \\ {\displaystyle v_0\left(0\right)\ge 0,v_0\left(1\right)\ge 0,v_0^{\prime \prime }\left(0\right)\le 0.}\hfill \end{array}\right.$$

Now, we give the following assumptions:

Assumption A1 

(${\mathit{H}}_{\mathit{1}}$). There exist $u_0$, $v_0\in E$ which are respectively lower and upper solutions of problem (2) and satisfying $u_0\le v_0$.

Assumption A2 

(${\mathit{H}}_{\mathit{2}}$). $f\in C\left(\right[0,1]\times \mathbb{R},\mathbb{R})$ and there exists $M>0$ such that

$$f(t,u)-f(t,v)\ge -M(u-v),t\in [0,1],u_0\left(t\right)\le v\le u\le v_0\left(t\right),$$

and M satisfies (5) and (16).

Theorem 1. 

Suppose that $\left(H_1\right)$ and $\left(H_2\right)$ hold. Then there exist monotone iterative sequences $\left\{u_n\left(t\right)\right\}$, $\left\{v_n\left(t\right)\right\}$ which converge uniformly on $[0,1]$ to the extremal solutions of problem (2) in the ordered interval

$$\Psi =\left\{\alpha \in E:u_0\left(t\right)\le \alpha \left(t\right)\le v_0\left(t\right),t\in [0,1]\right\}.$$

Proof. 

For given $u_0$ and $v_0$, we consider the following two linear fractional differential equations:

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }{u}_1\left(t\right)-M{u}_1\left(t\right)+f(t,u_0\left(t\right))+M{u}_0\left(t\right)=0,t\in (0,1),}\hfill \\ {\displaystyle u_1\left(0\right)=0,u_1\left(1\right)=0,u_1^{\prime \prime }\left(0\right)=0,}\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }{v}_1\left(t\right)-M{v}_1\left(t\right)+f(t,v_0\left(t\right))+M{v}_0\left(t\right)=0,t\in (0,1),}\hfill \\ {\displaystyle v_1\left(0\right)=0,v_1\left(1\right)=0,v_1^{\prime \prime }\left(0\right)=0.}\hfill \end{array}\right.$$

By Lemma 5, the solutions $u_1$ and $v_1$ are well defined. Let $\beta \left(t\right)=u_0\left(t\right)-u_1\left(t\right)$. According to the definition of the lower solution, it follows that

$$\begin{array}{cc}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\beta \left(t\right)}\hfill & {\displaystyle ={}^{C}D_{0^{+}}^{\nu }{u}_0\left(t\right)-{}^{C}D_{0^{+}}^{\nu }{u}_1\left(t\right)}\hfill \\ & {\displaystyle \ge -f(t,u_0\left(t\right))-M{u}_1\left(t\right)+f(t,u_0\left(t\right))+M{u}_0\left(t\right)}\hfill \\ & {\displaystyle =M\beta \left(t\right),}\hfill \end{array}$$

$$\beta \left(0\right)=u_0\left(0\right)-u_1\left(0\right)\le 0,\beta \left(1\right)=u_0\left(1\right)-u_1\left(1\right)\le 0,$$

and

$${\beta }^{\prime \prime }\left(0\right)=u_0^{\prime \prime }\left(0\right)-u_1^{\prime \prime }\left(0\right)\ge 0,$$

i.e.,

$$\left\{\begin{array}{c}\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\beta \left(t\right)-M\beta \left(t\right)\ge 0,t\in (0,1),}\hfill \end{array}\hfill \\ \beta \left(0\right)\le 0,\beta \left(1\right)\le 0,{\beta }^{\prime \prime }\left(0\right)\ge 0.\hfill \end{array}\right.$$

By Lemma 7, $\beta \left(t\right)\le 0$, $\forall t\in [0,1]$. Hence, $u_0\left(t\right)\le u_1\left(t\right)$. By the definition of the upper solution of problem (2), a similar argument yields $v_1\left(t\right)\le v_0\left(t\right)$$\forall t\in [0,1]$. Moreover, let $m\left(t\right)=v_1\left(t\right)-u_1\left(t\right)$. Form condition $\left(H_2\right)$, it follows that

$$\left\{\begin{array}{c}\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }m\left(t\right)-Mm\left(t\right)\ge 0,t\in (0,1),}\hfill \end{array}\hfill \\ m\left(0\right)\le 0,m\left(1\right)\le 0,m^{\prime \prime }\left(0\right)\ge 0.\hfill \end{array}\right.$$

Using Lemma 7 again, $m\left(t\right)\le 0$, $\forall t\in [0,1]$. Thus, we have established the relation $u_0\le u_1\le v_1\le v_0$.

Next, for any $u_{n-1}$, $v_{n-1}\in \Psi$, we may define two sequences $\left\{u_n\left(t\right)\right\}$, $\left\{v_n\left(t\right)\right\}$ satisfying the following equation:

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }{u}_{n+1}\left(t\right)-M{u}_{n+1}\left(t\right)+f(t,u_n\left(t\right))+M{u}_n\left(t\right)=0,t\in (0,1),}\hfill \\ {\displaystyle u_{n+1}\left(0\right)=0,u_{n+1}\left(1\right)=0,u_{n+1}^{\prime \prime }\left(0\right)=0,}\hfill \end{array}\right.$$

(19)

and

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }{v}_{n+1}\left(t\right)-M{v}_{n+1}\left(t\right)+f(t,v_n\left(t\right))+M{u}_v\left(t\right)=0,t\in (0,1),}\hfill \\ {\displaystyle v_{n+1}\left(0\right)=0,v_{n+1}\left(1\right)=0,v_{n+1}^{\prime \prime }\left(0\right)=0.}\hfill \end{array}\right.$$

By Lemma 5, $\left\{u_n\left(t\right)\right\}$, $\left\{v_n\left(t\right)\right\}$ are well defined.

Let $W_n\left(t\right)=u_n\left(t\right)-u_{n+1}\left(t\right)$. Then

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }{W}_{n+1}\left(t\right)-M{W}_{n+1}\left(t\right)\ge 0,t\in (0,1),}\hfill \\ {\displaystyle W_{n+1}\left(0\right)=0,W_{n+1}\left(1\right)=0,W_{n+1}^{\prime \prime }\left(0\right)=0.}\hfill \end{array}\right.$$

By Lemma 7, we obtain $W_n\left(t\right)\le 0$ for $t\in [0,1]$, i.e., $u_n\left(t\right)\le u_{n+1}\left(t\right)$ for $t\in [0,1]$. Similarly, we deduce that

$$u_0\le u_1\cdots u_n\le u_{n+1}\le v_{n+1}\cdots v_1\le v_0.$$

(20)

We show that $\left\{u_n\left(t\right)\right\}$ converges uniformly on $[0,1]$. In fact, it follows (20) that $\left\{u_n\right\}$ is uniformly bounded on $[0,1]$. By ($H_2$), there exists a constant $C>0$ such that

$$|f(t,u_n\left(t\right))+M{u}_n\left(t\right)|\le C,t\in [0,1],n=1,2,\dots .$$

(21)

By Lemma 5 and (7), the unique solution $u_{n+1}\left(t\right)$ to (19) is given by

$$u_{n+1}\left(t\right)={\int }_0^{1}H(t,s)(f(t,u_n\left(s\right))+M{u}_n\left(s\right))ds.$$

(22)

Combining this with (21), we conclude that $\left\{u_n\left(t\right)\right\}$ is equicontinuous. By the Arzela–Ascoli theorem, we obtain that $\underset{n\to \infty }{lim}{u}_n\left(t\right)=u^{*}\left(t\right)$ uniformly and monotonically on $[0,1]$. Furthermore, by (22), we conclude that $u^{*}$ is a solution of (2). Similarly, we can prove that $\underset{n\to \infty }{lim}{v}_n\left(t\right)=v^{*}\left(t\right)$ uniformly and monotonically on $[0,1]$, and that $v^{*}$ is a solution of (2).

In the end, we show that $u^{*}\left(t\right)$ and $v^{*}\left(t\right)$ are minimal and maximal solutions of (2) in $\Psi$, respectively. Let $x\left(t\right)\in E$ be any solution of (2) such that $u_0\left(t\right)\le x\left(t\right)\le v_0\left(t\right)$ for $t\in [0,1]$. Assume that for some positive integer n, $u_n\left(t\right)\le x\left(t\right)\le v_n\left(t\right)$ for $t\in [0,1]$. Let $\gamma \left(t\right)=u_{n+1}\left(t\right)-x\left(t\right)$. By $\left(H_2\right)$, we have

$$\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\nu }\gamma \left(t\right)-M\gamma \left(t\right){=}^C{D}_{0^{+}}^{\nu }(u_{n+1}\left(t\right)-x\left(t\right))-M(u_{n+1}\left(t\right)-x\left(t\right))}\hfill \\ {\displaystyle =-f(t,u_n\left(t\right))-M{u}_n\left(t\right)+M{u}_{n+1}\left(t\right)+f(t,x\left(t\right))-M{u}_{n+1}\left(t\right)+Mx\left(t\right)\ge 0.}\hfill \end{array}$$

Additionally, we have $\gamma \left(0\right)\le 0$, $\gamma \left(1\right)\le 0$ and $\gamma \prime \prime \left(0\right)\ge 0$. By Lemma 7, we obtain $\gamma \le 0$, that is, $u_{n+1}\left(t\right)\le x\left(t\right)$. Following a similar argument as above, we conclude that $x\left(t\right)\le v_{n+1}\left(t\right)$. Since $u_0\le x\le v_0$, it follows by induction that $u_n\le x\le v_n$ for all n. Taking the limit as $n\to \infty$, we have $u^{*}\le x\le v^{*}$. Thus the proof is concluded. □

4. Example

In this section, we provide an example to verify Theorem 1. We consider the nonlinear fractional differential equation as follows:

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\frac{5}{2}}\phi \left(t\right)+\frac{3}{4}\left[\frac{1}{6}{\left(t-\phi \left(t\right)\right)}^3+t^{2}sin\frac{\phi \left(t\right)}{4}\right]=0,t\in (0,1),}\hfill \\ {\displaystyle \phi \left(0\right)=\phi \left(1\right)=\phi \prime \prime \left(0\right)=0.}\hfill \end{array}\right.$$

(23)

Here, $\nu ={\displaystyle \frac{5}{2}}$ and $f(t,u)={\displaystyle \frac{3}{4}}\left[{\displaystyle \frac{1}{6}}{(t-u)}^3+t^{2}sin{\displaystyle \frac{u}{4}}\right]={\displaystyle \frac{1}{8}}{(t-u)}^3+{\displaystyle \frac{3t^2}{4}}sin{\displaystyle \frac{u}{4}}$.

Taking $u_0\left(t\right)=0$, $v_0=3t$, we have

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\frac{5}{2}}{u}_0+\frac{3}{4}\left[\frac{1}{6}{\left(t-0\right)}^3+t^{2}sin\frac{0}{4}\right]=0+\frac{3}{4}\left(0+\frac{1}{6}{t}^3\right)\ge 0,t\in (0,1),}\hfill \\ {\displaystyle u_0\left(0\right)=u_0\left(1\right)=u_0^{\prime \prime }\left(0\right)=0,}\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\displaystyle {}^{C}D_{0^{+}}^{\frac{5}{2}}{v}_0+\frac{3}{4}\left[\frac{1}{6}{\left(t-3t\right)}^3+t^{2}sin\frac{3t}{4}\right]=0+\frac{3}{4}\left[-\frac{4}{3}{t}^3+t^{2}sin\frac{3t}{4}\right]}\hfill \\ {\displaystyle \le \frac{3}{4}\left[-\frac{4}{3}{t}^3+t^2\frac{3t}{4}\right]=-\frac{7}{16}{t}^3\le 0,}\hfill \\ {\displaystyle v_0\left(0\right)=v_0^{\prime \prime }\left(0\right)=0,v_0\left(1\right)=3\ge 0,}\hfill \end{array}\right.$$

which shows that $u_0\left(t\right)$ and $v_0\left(t\right)$ are the lower and upper solution of (23).

Additionally, considering the function $f(t,u)$, we calculate its partial derivative with respect to u. Noting that $cos{\displaystyle \frac{u}{4}}\ge 0$ for $u\in [0,3]$, we have

$$\begin{array}{cc}{\displaystyle \frac{\partial f(t,u)}{\partial u}}\hfill & {\displaystyle =-\frac{3}{8}{(t-u)}^2+\frac{3t^2}{16}cos\frac{u}{4}\ge -\frac{3}{8}{(t-u)}^2}\hfill \\ & {\displaystyle \ge -\frac{3}{8}{(t-3t)}^2=-\frac{3}{2}{t}^2\ge -\frac{3}{2},0\le u\le 3t,t\in [0,1].}\hfill \end{array}$$

Consequently, an application of the mean value theorem yields

$$f\left(t,u\right)-f\left(t,v\right)\ge -{\displaystyle \frac{3}{2}}\left(u-v\right),0\le v\le u\le 3t.$$

We therefore select $M={\displaystyle \frac{3}{2}}$. With this choice, it is readily verified that

$$\begin{array}{cc}\hfill M& ={\displaystyle \frac{3}{2}}<min\left\{{\displaystyle \frac{\Gamma (\nu +1)}{\nu -1}},(\nu -2)\Gamma (\nu +1)\right\}={\displaystyle \frac{15}{16}}\sqrt{\pi }\approx 1.6617,\hfill \\ \hfill M& ={\displaystyle \frac{3}{2}}<{\displaystyle \frac{{\nu }^{\frac{\nu }{\nu -1}}\Gamma (\nu +1)}{\nu -1}}={\displaystyle \frac{5}{4}}{\left({\displaystyle \frac{5}{2}}\right)}^{{\displaystyle \frac{5}{3}}}\sqrt{\pi }\approx 10.2006.\hfill \end{array}$$

Thus, all the inequalities specified in assumption $\left(H_2\right)$ are strictly satisfied.

In conclusion, the conditions of Theorem 1 are satisfied. Thus, there exist sequences $u_n\left(t\right)$, $v_n\left(t\right)$ that converge uniformly to the minimal and maximal solutions of (23) in $[u_0,v_0]$, respectively.

5. Conclusions

This paper investigated the existence of extremal solutions for a class of Caputo fractional differential equations of order $\nu \in (2,3]$. The key contribution of this work is the establishment of two novel comparison principles. By coupling comparison principles with the classical monotone iterative technique and the method of upper and lower solutions, we successfully circumvented the monotonicity constraints usually required for the nonlinear terms.

Furthermore, a constructive approach for explicitly deriving these extremal solutions has been provided. The practical applicability of our theoretical framework is verified through an illustrative example. We expect that the findings of this paper will not only broaden the scope of monotone methods in treating non-monotone fractional boundary value problems but also serve as a solid foundation for studying more complex fractional order systems.